You are given an m x n binary matrix grid. An island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.
The area of an island is the number of cells with a value 1 in the island.
Return the maximum area of an island in grid. If there is no island, return 0.
Example 1:
Input: grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]
Output: 6
Explanation: The answer is not 11, because the island must be connected 4-directionally.
Example 2:
Input: grid = [[0,0,0,0,0,0,0,0]]
Output: 0
Constraints:
- m == grid.length
- n == grid[i].length
- 1 <= m, n <= 50
- grid[i][j] is either 0 or 1.
class Solution {
public int maxAreaOfIsland(int[][] grid) {
int max_area = 0;
for(int i = 0; i < grid.length; i++)
for(int j = 0; j < grid[0].length; j++)
if(grid[i][j] == 1)max_area = Math.max(max_area, AreaOfIsland(grid, i, j));
return max_area;
}
public int AreaOfIsland(int[][] grid, int i, int j){
if( i >= 0 && i < grid.length && j >= 0 && j < grid[0].length && grid[i][j] == 1){
grid[i][j] = 0;
return 1 + AreaOfIsland(grid, i+1, j) + AreaOfIsland(grid, i-1, j) + AreaOfIsland(grid, i, j-1) + AreaOfIsland(grid, i, j+1);
}
return 0;
}
}